Быдло математику прогуливает, познания гопоты ограничиваются только тем, как бы отнять и поделить. Так что да, тут только культурная братва, остальным делать особо нечего. 😉
You might lose a minus sign in that case. I would draw a circle in the complex plane and make sqrt(i) and sqrt(-i) into two vectors. I could solve this in my head without pen and paper.
Each root individually has 2 solutions, so there should be 4 solutions. If you map the solutions of root(i) and root(-i) on the complex plain (recall that by taking a root of a complex number you half the phase angle, but you can do this in either direction giving you 2 solutions). And if you then do the addition you'll easily see that all 4 solutions are in a different axis on the complex plain, and that they should be root(2), -root(2), i*root(2) and -i*root(2).
Iagree, if you look at it geometrically, this pops up easily. Just think of i=e^(ipi/2) and the squareroot as dividing that angle into half, so one solution is sqrt(i)=e^(ipi/4) , the other sqrt(i)=e^(i5pi/4). Same with -i, which leads to sqrt(i)=e^(i3pi/4) and sqrt(i)=e^(i7pi/4). Now add them pairwise like vectors, which should lead to all 4 solutions sqrt2, i×sqrt2, -sqrt2 and -i×sqrt2. I would never recommend calculations with complex numbers in only one kind of expression, but switching back and forth and sideways between sum form, exponential form and as geometrical object. By the way, seeing complex numbers as rotation also helps, when multiplied.
@@ralfp8844 Exactly my experience. Algebraic manipulation is the most powerful, but easy to make an error. Oftentimes, geometric representations really helps understanding the problem and breaking it down into easier sub-problems.
Могу сослаться на решение этой задачи в "A Wonderful Math", указанном здесь в первом комментарии. Там представлено решение без возведeния в четвертую степень и тоже только два корня +/- корень из 2 ... .
that would be true, except the radical sign represents the positive solution. if you want both solutions you must write +-✓ . So for this question it is asking for only the positive roots, and so there is 1 solution. However if the question was written as +-(✓i +-✓-i) then there would be four solutions as you say.
I use polar form of complex number for general solution: i = e^(2m+1/2)πi sqrt(i) = e^(m+1/4)πi -i = e^(2n-1/2)πi sqrt(i) = e^(n-1/4)πi sqrt(i) + sqrt(-i) = e^(m+1/4)πi + e^(n-1/4)πi = z Now we have 4 cases to deal with (m,n) is (even, even): z = e^(π/4)i + e^-(π/4)i = 1/√2 + (1/√2)i + (1/√2 - (1/√2)i) = √2 (m,n) is (even, odd): z = e^(π/4)i + e^(3π/4)i = 1/√2 + (1/√2)i + (-1/√2 + (1/√2)i) = (√2)i (m,n) is (odd, even): z = e^(-3π/4)i + e^-(π/4)i = -1/√2 - (1/√2)i + (1/√2 - (1/√2)i) = -(√2)i (m,n) is (odd, odd): z = e^(-3π/4)i + e^(3π/4)i = -1/√2 - (1/√2)i + (-1/√2 + (1/√2)i) = -√2 All four solutions have modulus √2 and either purely real or purely imaginary. We can easily generalize all of them. sqrt(i) + sqrt(-i) = √2 e^(kπ/2)i where k is any integer.
Some people argue that sqrt means "principal values only" I don't know if it is true, if yes, you have to adjust all angles within (-π,π] √2 is the principal value in this case.
Actually there are 4 answers, in general ±((√(2)+i√(2))/2)±((√(2)-i√(2))/2) The main answer is √(2) But there is also -√(2) and i√(2) And also -i√(2) ±√(2) and ±i√(2)
I won't say he "lost" the ± but he made an error when he factored out the ±1/sqrt(2). There is no such thing as a ±1/sqrt(2). It is plus OR minus. Not both. That means there are 4 solutions: a ++ version, a +- version, a -+ version and a - -- version. On the other hand, I thought his solution was clever. Yeah, there are other ways (square them, convert to exponential functions, etc) but this was fun. Other than the factorization error.
@@flowingafterglow629 You are of course correct. This may be written as the two expressions ± a (b + c), which is his two solutions and ± a (b - c), which is the missing two. Still 4 solutions, but written in two expressions.
@@flowingafterglow629 I mean ”lost”. That is: not ”lost”, but ” ”lost” ”.😄 I thought there are such an English style to say something like ”+/-sqrt(2)”, just because the solution was too clever to believe it contains such a plain error.
@@flowingafterglow629 The square root radical is defined as being the principal branch **only**. Thus the correct answer of the problem as posted is single valued at sqrt(2) * i evaluated as (1 + i)/sqrt(2) + (1 - i)/sqrt(2) = (2 + 0i/sqrt(2) = sqrt(2). Above corrected for a brain-fart error in the original. The "principal branch" defined for the square root radical is obtained by making a "branch cut" along the negative real axis. This is effected by restricting the value of theta for the original expression (in exponential form) under the radical to being between - pi and + pi.
@@pietergeerkens6324 What is the "principal (sic) branch" of a complex number? Is exp(i*pi/2) any more "principle" than exp(3*i*pi*2)? Is there a rule that the complex squareroot of a number is the one in the top half of the complex plane?
There's only one answer technically if you consider how square roots are defined in the context of complex numbers, the solution with positive real part is chosen while removing the radical... If real part is zero, the one with positive imaginary part is chosen. By chosen, I mean, it becomes the principal square roots. In general, any given complex number can be expressed in infinitely many ways in the form r•e^(i(x+2nπ)) where n is any integer. So when we apply say square root, the number becomes r•e^(i(x/2+nπ)). We choose n such that x/2 + nπ is an acute angle, (which if not possible, our next preference is obtuse angle or the negative version of acute angle). That's about it... So there's only one square root of a complex number... √i + √(-i) = √(e^(iπ/2)) + √(e^(-iπ/2)) = e^(iπ/4) + e^(-iπ/4) = 2cos(π/4) = √2 That's the only solution if we consider only the *principal square roots* of the radicals involved.
In practice the "principle" square root is never used. If you are already working with complex numbers, in almost all fields of science you are interested in all roots.
@@crossiqu if we use the principle cube root of -i, we get that it becomes i. If we instead try to find all solutions to x^3=-i, we have x=i, and x=-i/2±√(3)/2.
-sqrt(2), i*sqrt(2), -I*sqrt(2) are also solutions: it depends on which branch of square root you use. Imagine solutions are when you use different branches for the root.
i solved it "geometrically": visualize i and minus i as arrows in the plain of complex numbers. i has ankle 90 degrees, -i has ankle 270 degrees. length of both arrows is 1. interprete the operation sqrt() as division of ankle by 2. so you get two arrows, that represent sqrt(i) and sqrt(-i). add these two arrows as known from vector addition. you get a triangle which leads you to the solution sqrt(2)*i, if you apply some basic trigonometry on it. (in triangle, use: sin(45°)=0.5sqrt(2))
Your method is indeed the polar form. If you talk about all possible solutions, there are 3 more. If you want principal value only, your solution is not. You need to adjust all initial polar angles to [-π, π] before doing anything.
@@mokouf3 "Your method is indeed the polar form" yes. "If you want principal value only...." i have heard of principal forms, but didn't understand them properly yet. i'll think about it the next days. how can there be more than one solution to a summation? how about the contradiction that i showed in the proof above? the contradiction should be wrong if you're right. where exactly is it wrong?
@@KarlHeinzSpock There's multiple solution because i has multiple angles. It doesn't have to have a angle of 90. It can also have a angle of 450, or 810, or any angle of the form 360k+90. The polar coordinates (90,1), (450,1), (810,1) all give you the same point on the plane.
@@hrtz9796 all angles of the form 360k+90 deliver only one and always the same number, i. but sqrt(i) has indeed two solutions, as well as sqrt(-i) -->the sum sqrt(i)+sqrt(-i) "includes" four different sums, thus has four different solutions.
As some astute commenters have pointed out, there are four solutions. The video limits solutions to the real numbers, but the problem involves complex numbers so I think they’re fair game for the solution. You can derive SQRT(i) by setting it equal to +/- (a + bi) and doing the algebra, but that isn’t necessary. A moment’s thought should bring you to the conclusion that i multiplied by the real roots gives valid solutions because squaring i just flips the sign from +/- to -/+, and both signs are already part of the solution.
Technically, that'll give you the numeric value of the result, but it isn't hard to convince ourselves that the correct result must be the positive root.
@@jensknudsen4222 both roots are ok. Maths is a world by itself, and while the “positive” root is sometimes preferred for Analysis and Physiscs, the convention of the positive root being the “correct” one is not as universaly accepted as most people think. If you find this a silly thing, try to take the square root of x=a^2+b^2, and then try to decide wich one is the positive or the negative root.
Such a procedure is always prone to introducing extraneous roots. Eg let x=-1. If you square and square root you get two roots, 1 and -1. Only the second is correct in the setup.
First of all, the square root (√ ) of complex numbers are defined only by the polar form such that the argument becomes halved and the old magnitude is the square of the new magnitude. Eg √(z)=√(r)e^(it/2) given that z=re^(it). This means that the method you use might give the wrong result, but it seems like it does not. Second, even for real numbers, √(x)>0, so even though the solution for x^2=a has two solutions, √(x) only has one solution for all x>0. Thirdly, the function √ also only have one solution for the complex domain, where we may only choose the polar representation that gives us 0
First, the complex square root is most commonly defined only for complex numbers whose argument lies in the semi-closed interval (-π, +π] -- in other words, excluding -π, but including +π. Then, halving the argument ensures that the principal square root thus obtained is consistent with real square roots and remains on the same branch for complex numbers. That's why the method works. Second, √(x)>0 only remains valid for complex numbers if you restrict the domain of x in the way I outlined above. Otherwise, consider the square root of x that your definition produces when x = e^(2πi). Thirdly, your definition "the function √ also only have one solution for the complex domain, where we may only choose the polar representation that gives us 0
The sqrt bracket operator denotes the *principal" square root, hence, no need for considering the negative roots. In other words, it should have been reduced to sqrt(2)*(1 + i + 1 - i) = sqrt(2)*2 = sqrt(2). No need for +/-. If you *do* want to consider non-principal square roots, then you also need to consider the cases where you have opposite selections for the +/- on either root. Hence, sqrt(2)*((1+i)-(1-i)) = sqrt(2)*2*i = sqrt(2)*i. Similarly, another non-principal square root would be -sqrt(2)*i.
Sorry to nitpick on your solution, but I think you're leaving out an important (though technical) detail. The square root symbol can actually mean two different things: 1. The roots of z^2 = a. For complex a != 0 there are always two roots, one being the negative value of the other. Viewed as a function, this version of the square root is multivalued. 2. The principal value, which is defined as the one of the two roots that has a positive real value (or +i in case the real values are zero). Note that this version is a regular (single valued) function. Meaning (1) is rarely used, and it's normally pretty safe to assume that the square root symbol refers to the principal value unless otherwise specified. Assuming this to be the case (that is, both square roots in the given problem both refer to the principal value), the expression only has one value: sqrt(i) + sqrt(-i) = (sqrt(2)/2)(1 + i) + (sqrt(2)/2)(1 - i) = sqrt(2). If, on the other hand, if we let the square root symbol denote both roots of the quadratic, then each of the square roots in the original expression has TWO values. Each choice is independent of the other, so we end up with FOUR values: sqrt(i) + sqrt(-i) = +/-(sqrt(2)/2)(1 + i) +/- (sqrt(2))(1 - i) where the two +/- operators are independent. Evaluating this all four possible ways yields four distinct values: sqrt(2), -sqrt(2), sqrt(2) i, -sqrt(2) i.
That’s not the only way roots of polynomials work. In fact, the answer depends on the splitting field you want to decompose the polynomial in. If i’m working on Q(sqrt(i);sqrt(-i)) the answer is four-valued, as you state. If i’m working on R(sqrt(i);sqrt(-i)), the answer is +-sqrt(2), as in the video (this one is actually the most common viewpoint), however, if i’m working in C(sqrt(i);sqrt(-i)), then the answer is still sqrt(i)+sqrt(-i). So there are, in general many, many ways to actually “solve” this problem.
@simonflavioibanez7715 Haha... Yes, that's why mathematicians will never be bored. Just when you think you're done, some generalization comes swooping in and knocks over your dominoes.
I just visualized the vector for i and halved the argument, and then the same for -i. The imaginary parts cancel, and the real parts add, and there’s your answer.
But did you take the argument for -i to be 3π/2 or -π/2 ? Both are valid arguments for -i, but you get two different complex numbers when you halve the argument. That's a result of the fact that all numbers (except zero), real or complex, have two distinct square roots that are the negative of each other. Had you picked the other argument, you'd find the real parts cancel and the imaginary parts add :o
@@RexxSchneider Every non-zero number does indeed have two square roots, but the radical symbol refers explicitly to the principal square root, which I believe to be defined as the one obtained by starting with -pi < arg(z)
@@MikeGranby If you read the other threads here, you'll find that many commentators don't define the √ to indicate the principal square root when dealing with complex numbers. Others accept that but use a different choice of branch cut such as 0
@@RexxSchneider I can see how one might use a different branch cut, although the one I quoted is the one I was taught, but I think allowing the square radical to be multivalued for is a step too far…
a sum of numbers must not have two different values. only one of the solutions +-sqrt(2)i is correct. proof: for example: 0=0 sqrt(i)+sqrt(-i)=sqrt(i)+sqrt(-i) sqrt(2)i=-sqrt(2)i 2sqrt(2)i=0 this is obviously a contradiction in itself. only the solution +sqrt(2)i is correct.
You missed 1 possible answer: i = 1*cis(90) ==> sqrt(i) = sqrt(1)*cis(90/2 + 360/2 * k), k = 0, 1 so 2 possible answers for sqrt(i): a= 1*cis(45) b= 1*cis(225) -i = 1*cis(270) ==> sqrt(-i) = sqrt(1)*cis(270/2 + 360/2 * k), k = 0, 1 so 2 possible answers for sqrt(-i): c= 1*cis(135) d= 1*cis(315) Now we can look at all the combinations: a+c = 0 a+d = sqrt(2) b+c = -sqrt(2) b+d = 0 This is all very easy if you look at it in polar form, and addition is just vector addition.
I find a few problems in the calculation. Sum of two unique values should result in one answer. If your answer is back substituted sqrt(i)+sqrt(-i)=+/-sqrt2. Squaring as is i + 2 sort(i)sqrt(-i) - i =2 2sqrt(i)sqrt(i)sqrt(-1)=2 2i.i=2 2i^2=2 -2=2 so your answer doesn’t satisfy original equation. isqrt2 would be the correct answer. When you back substitute LHS will simplify the same and the RHS will have an extra -i^2 term making it -2=-2.
By the definition of the the sqrt sign : sqrt(z) is THE principal square root of z, which is unique. For example, sqrt(1)=1 not -1, and sqrt(i)=1/sqrt(2) + i*sqrt(2), Though the equation z^2 = i has two solutions. qrt(-i)=1/sqrt(2) - i*sqrt(2),
The simple and normal solution is by squaring the original equation (gives)... i cancel - i and let 2 multiplied by squared (i) then returning rooting the equation gives =(+&_)root 2 Thanks sir.
Painful. Just use and exp(ix) = Cosx + iSinx, and don't forget the +- roots. Then you combine +/-exp(iPi/4) and +/-exp(-iPi/4). You get 4 combinations leading to 2 values: root2 or -root2.
It's not an equation. It is a simple calculation involving functions, which by definition returns only one value. So there is ONLY ONE VALID ANSWER : +V2
@@tontonbeber4555 If you just talk about principal values sqrt(i) = e^(π/4)i sqrt(-i)= e^-(π/4)i Their sum is indeed √2 But, unlike real functions, a complex function tend to be multi-valued.
@@boblewis5762 If you talk about principal values only, this one is not used. If you talk about multi valued version, my general solution included that.
The sqrt symbol means principle value only so don’t use +/-. Also, you made it needle complicated. It’s just 2* Re( sqrt( i) ). And i is exp( i • pi/2 ) so we get 2* Re( exp( i • pi/4 ) = 2* 1/ sqrt( 2 ) = sqrt( 2 ). Ans: sqrt( 2 ) without +/-.
Where did you get the "just 2* Re( sqrt( i) )" from? Did you work backwards from what you thought the answer was? You could also have said that i = exp( i * -3pi/2) so we get 2* Re( exp( i • -3pi/4 ) = 2* (-1/ sqrt( 2)) = -sqrt(2) :o
Just as easy, y = sqr(i)+sqr(-i). y^2=(sqr(i))^2 + 2sqr(i*-i) + (sqr(-i))^2 = i +2 sqr(1) - i = +/- 2 so y = sqr(y^2) = sqr(+/- 2) = +/- sqr(2), +/- i sqr(2) Nice video but seemed a bit confusing. A beginner would have no idea how to make those clever changes.
Once you write the symbol "√" the choice of sign has already been chosen by you. √4 = 2 not ±2. If you want ±2 then you would need to write ±√4. The square root function is a function. You can only have one output. It maps C to {z in C | Re(z)≥0}. So √i = (1+i)/√2 and not - (1+i)/√2. Similarly √(-i) = (1-i)/√2 . Their sum is √2.
@@crossiqu What are you talking about? It is a function on the complex plane holomorphic everywhere in C excluding the negative real numbers and 0. The question of this video is about square roots of i and -i, both of which are no where near the negative real axis. Hell, you can take the derivative of it at i and -i. The principal square root, which is universally understood when the √ symbol is written, is well defined.
@@DrR0BERT It's simple. Square root of complex numbers generalized, is not a function in the sense its not return a single value. But you're right for real numbers, and IF you're talking about the PRINCIPAL square root of a complex number. So, OK, the principal square root function is thus defined using the nonpositive real axis as a branch cut. But the written symbol can express both. Its an historic topic regarding n nth ROOTS for nth polinomia. (cube roots and beyond.What are the cube roots of -i?) So the formula it's a kind of ambigous
@@crossiqu What are the cube roots of -i? There are three of them. That's not the issue. Nor is the issue that both i and -i have two square roots. The issue is when you use the symbol √ you are identifying only one of them, namely the principal square root. If you want to reference the non-principal square root, you would use -√ . Let me make this simple. The two questions: "What is √(-i)?" and "What are the square roots of i?" do not have the exact same answer. The first one is a set of one number, and the second is a set of two.
@@DrR0BERT You may find it helpful to understand that using the surd symbol √ to mean the principal square root of a number has not always been the convention. You will find that different audiences of different nationalities and different text books may utilise conventions in a way that differs from yours. Admittedly, you've probably got to examine text books written in English that are over 100 years old to see √ used to mean both square roots, and I agree that the majority of serious mathematicians working in English today would expect the symbol √ to denote the principal square root, but I don't think it's helpful to simply dismiss alternative interpretations as invalid.
Let the single letter p be pi, sqrt(i) + sqrt(-i) = cis(p/4)+cis(-p/4), vector1= 45 degree (positive x axis) with a length of 1 from origin +vector2= -45 degree with a length of 1 from origin, give you the resultant vector of sqrt(2) length lying on the positive X axis, so answer is real number sqrt(2) [Right angle triangle]
There is only one value for this, the square root is not a multi-valued function by definition, and your insertion of a +/- is unambiguously wrong. This solution is not only overly long (compared to using the exponential definition) but also incorrect.
I agree with sqrt(I)=(1+i)/sqrt2 sqrt(-i)=sqrt(-1)*sqrt(i)=i(1+i)/sqrt2=(-1+i)/sqrt2 So sqrt(i)+sqrt(-i)=2i/sqrt2 Using polar form you get the same result. This answer satisfies original equation.
Assuming that you are taking the surd symbol √ to indicate the principal square root of the complex number, we proceed like this: Write the complex number in polar form, so i = e^(πi/2). That has two square roots, e^(πi/4) and e^(πi/4 + πi). The principal square root of a complex number r.e^(iθ) whose argument (θ) satisfies -π < θ
i looked up some basics. let C be the set of all complex numbers. the operation + of two complex numbers is then defined as a function '+': (CxC)-->C. this is part of a group theory axiom. since '+' is a function, the sum x+y of two complex numbers must have exactly one solution. so only +i*sqrt(2) is a solution. proove me wrong, if you know something that i don't know or that i misunderstood, since i'm not completely shure about what i said above. it's not about who knows best, it's about learning something.
i'm on the trail of my own misunderstandings: sqrt(i) is not only one number, there are two numbers coming out of sqrt(i). it can have the angles 45° and 225° sqrt(-i) can result in an angle of 135° and an angle of 315° so sqrt(i)+ sqrt(-i) represent four different additions with four differe values.....
To do it the way you did it, you kind of have to anticipate the answer. Why not use 3/3 or 4/4 instead of 2/2? Using 2/2 allows you to get to (1+i)^2 and (1-i)^2 and solve it the way you did. The answers in the comments that did it geometrically or by squaring both sides showed me why sqrt(2) is correct.
Just use i and -I exponential notation you get e^pi/4+e-pi/4 by simply fine using the square roots and the properties of exponential but this is 2cos(pi/4) using Euler’s trigonometric identify which is equal to sqrt(2) finally
+/- sqrt(a) +/- sqrt(b) Can also be: sqrt(a) - sqrt(b) & -sqrt(a) +sqrt(b) i^(1/2) + (-i)^(1/2) = e^(i x pi / 2 + 2 x pi x m)^(1 / 2) + e^((-i) x pi / 2 + 2 x pi x n)^(1 / 2) = e^(i x pi / 4 + pi x m) + e^((-i) x pi / 4 + pi x n) = (e^pi)^m x e^(i x pi / 4) + (e^pi)^n x e^(-i x pi / 4) Now without the m: e^(i x pi / 4) =sqrt(2)/2 + sqrt(2)/2 & without the n: e^(-i x pi / 4) =sqrt(2)/2 - i x sqrt(2)/2 If m=n=+/- 2 then you can to this. But if m= 2 & n = -2 you have another solution. And with m=-2 & n=2 yet another. It's not difficult to see there are other solutions . For all odd m & n.
Wouldn't it be easier to square the whole thing, then +i and -i cancel out and in the "2*sqrt(i * -i)" part the square root evaluates to 1, leaving just 2. Take a square root of that and add +/- in front of it.
In this kind of a problem you have to add the principal square roots. Principal square root of -i is (i-1)/sqrt2 So your answer would be isqrt2. If you used polar form the problem wouldn’t be there. I=e^PI/2 so sqrt (I)=e^PI/4 = (1+i)/sqrt2 -I=e^3PI/2 so sqrt(-I)=e^3PI/4=(-1+i)/sqrt2 When you add the two you get the same answer I quoted.
I have simple solution for above equation, may be required correction: Let √i + √-i = x squaring on both side, (√i + √-i)² = x² Solving LHS, (√i + √-i)² = (√i)² + 2(√i)(√-i) + (√-i)² = i + 2√(i)(-i) -i =2√-(i²) =2√-(-1) =±2 After simplifying the left-hand side, we have (√i + √-i)² = ±2.
Very lengthy process followed. Just take square of the whole expression and expand. Then if it is equal to square of 'x', then retake the square root of the simplified result to get value of 'x' i.e. the given expression.
It is allowed. Any negative imaginary number can be written as r * e^(-π/2) where r is a positive real number. Its principal square root is normally defined to be √r * e^(-π/4), taking the positive value of the square root of r. Its other square root is √r * e^(3π/4).
Geometrically, you end up with 4 vectors 90 deg apart and splitting each quadrant in half. Adding vectors geometrically, you are finding the diagonals of 4 unit squares each of which lie on the R axis or the I axis. Of course, the diagonal of a unit square = sqrt(2).
If you're familiar with the properties of complex numbers this problem can be solved by inspection. By writing i in polar form one finds sqrt(i) + sqrt(-i) = +/- 2·cos(π/4) = +/- sqrt(2) ◼ No complex algebra required.
You've assumed that the ± i.sin(π/4) cancels. But the ± is independent for the case of sqrt(i) and sqrt(-i), so we find that in the four resulting cases, two of them cancel the imaginary parts, giving ±sqrt(2), but the other two cases cancel the real parts, giving ±i.sqrt(2).
@@RexxSchneider Pardon me for pointing it out, but you're wrong. i = e^(π i/2 + 2 π k i), where k is an integer. Therefore, sqrt(i) = e^(π i/4 + π k i). It's obvious then that sqrt(- i) is the conjugate of sqrt(i), and k is in the set {0,1}. When k = 0, sqrt(i) = cos(π/4) + i sin(pi/4); sqrt(-i) = cos(pi/4) - i sin(π/4). Summing both expressions together yields 2·cos(π/4). When k = 1, sqrt(i) = - cos(π/4) - i sin(π/4); likewise, sqrt(-i) = - cos(π/4) + i sin(pi/4). Again summing these two expressions yields - 2·cos(π/4). These are the only values of the original expression sqrt(i) + sqrt(-i), as assigning k any value greater than 1 or less than 0 causes the cycle to repeat, yielding the same two possible answers: +/- 2·cos(π/4) ◼
@@johnnolen8338 I'm also sorry to point it out, but you're wrong. The k in the polar form you write for i and the k in the polar form you write for -i are independent. You could call them k1 and k2 if you wish, and you're quite right that they both lie in the reduced set { 0, 1 }. But you've missed the case where k1 = 0 and k2 = 1 and the case where k1 = 1 and k2 = 0. Try those out and see which parts cancel out. Did you find the other two possibilities, now?
@@RexxSchneider I'm not going to debate this; k is an integer. As such you only have to include it once when defining i in polar form. -i is the complex conjugate of i. Therefore the k's are not independent. They have the same absolute value but opposite signs. Whenever you add a complex number to its complex conjugate the sum is always twice the real part of the complex number. The imaginary parts cancel because they have opposite signs. If k is even you get the positive value else if k is odd,you get the negative value. There are no other possibilities. Don't believe me? Plot it on the unit circle for yourself. But in the meantime stop playing stump the dummy with people who are way more experienced than you.
@@johnnolen8338 There's nothing to debate. If you can't see that there are two numbers, and each of them can have *any* integer in its polar representation., then you're too stupid to bother explaining things to. Your k's are clearly independent. If we were evaluating √(ai) + √(bi) would you still insist that √(a.e^πi(2m + 1/2) + √(b.e^πi(2n + 1/2) had to have n=m? Of course not, there's no such restriction. Then ask yourself what's magical about a=1; b=-1 that suddenly forces n to be equal to m? You state "Whenever you add a complex number to its complex conjugate the sum is always twice the real part of the complex number." Yes, of course. We know that (a+ib) + (a-ib) = 2a. But we're dealing with √(a+ib) and √(a-ib) and we can't are just call them "conjugates". The square root of i has two distinct values, A = (1+i)/√2 and B = (-1-i)/√2. The square root if -i has two other distinct values, C = (1-i)/√2 and D = (-1+i)/√2. You get four distinct possible sums when you add them together, A+C, A+D, B+C, B+D. You can't just decide to ignore two of those. Work it out and see how much you've embarrassed yourself. Of course I don't believe you. I'm a 71-year-old Cambridge graduate who spent a lifetime teaching maths, and you're an ignorant child who is unlikely to ever match my experience if you live another 100 years.
You write i^2-2.1.i+i^2. as (1-i)^2. and then use that ⎷(1-i)^2. equals 1-i What if you write i^2-2.1.i+i^2. as (i-1)^2. Then you would have found ⎷(i-1^2. equals i-1. Or not? The problem is that when you have ⎷(1-i)^2 and replace it by 1-i. you have to find out first if 1-i or i-1 is the principal root of (1-i)^2. Because ⎷ stand for the principal root of a value. And then in the situation of (i-1)^2 we also have ⎷(i-1^2. equals 1-i. Just using the fact that ⎷(1-i)^2. equals 1-i and not i-1 without an explanation makes the solution incomplete.
Two answers...that is great, but n general a square root of a complex number remains multi valued so in that sense the question is "worded" in a non conform way. It would be better to call the given expression x and then through squaring obtain a polynomial equation and then find all the roots. Equally, the assumption i = sqrt(-1) is incorrect, it should be "i is the imaginary unit for which its square is -1"
As I - not always, but often - do when I watch math videos, I stopped the video before watching the solution, took paper and pen and tried it on my own. I got the result i*sqrt(2). Then I continued watching the video to compare. After that I was surprised to see sqrt(2). At first I thought I had done something wrong, but after reading the comments I was confirmed, that my result was correct as well. So, all in all, there are 4 different possible solutions. - Interesting question, that was fun!
While that works in this case I'm not sure that is valid in general. For anything that summed to a number with negative real part it would be wrong (just (-0.5)+(-0.5) would fail it giving +1) .
I humbly disagree with everyone here asserting that there are four "solutions" to this problem. As this is presented, we are not being asked to solve for anything. We are simply asked to simplify the given expression. If we were instead given the expression (√4 + √9), one would simply evaluate the principle root of each term, and give the answer 5. At least as I was always taught, -5, -1, and 1 are not "alternative" or further correct answers. However, if we had been given two quadratic equations, x=√(4) and y=√(9), and asked to evaluate the sum x+y, then our solution set would consist of all four elements. The fact that this problem involves complex numbers seems to me irrelevant to this general point. Here we are only evaluating an expression, so we sum the two principle roots. Hence, √2*i At least that's what my math teachers from middle school through college told me. 🤷
to me (french, master degree in electronics) the sqare root function is only defined from R+ to R+. There is no sqrt(i) in my math. Nowhere to be seen. Only complex number x so that x^2=i (or any other complex number and any number of powers of x). I had never seen
I think what many commentators are saying is that they can see how √i + √(-i) could be evaluated to yield four different values. If you have been taught -- as the majority have -- that functions have only one value, that therefore we have to choose a "principal value" from the two possible square roots of any number, and that the surd symbol √ denotes the principal square root, then you will only find one value for √i + √(-i), namely √2. Nevertheless, I think from the volume of commentators who take a different approach, namely that √ can represent either the principal square root or its negative, that your understanding does not encompass the entire field of possibilities, and it is instructive to understand other views.
As far as I know, this term is unsovable, because negative numbers don´t have a square root. If "i" is positive, then "-i" must be negative and vice versa. If I am wrong, please tell me why. Thanks a lot.
your answer is correct, if you refer to the set of real numbers, those "normal" numbers known from school maths. the given math problem refers to a extended set of numbers, named complex numbers. complex numbers allow square roots of negative numbers. if you want to understan it, read first about complex numbers and their properties. i can't explain it here, it is too much stuff.
From the polar form, sqrt(i) = +/- [sqrt(2)/2 + i sqrt(2)/2] sqrt(-i) = +/- [sqrt(2)/2 - i sqrt(2)/2] sqrt(i) + sqrt(-i) = +/- sqrt(2). It was fun seing him do it the hard way 🤣
Если обе стороны уравнения возвести в квадрат, x²=(√i+√(-i))². Тогда: ▫x²=i+2√i√(-i)-i; ▫x²=2√i√(-i); ▫√a√b=√(ab); ▫√(i*(-i))=√(-i²); ▫i²=-1; ▫-(-1)=1; ▫√1=1; ▫x²=2*1; ▫x²=2; ▫x=±√2.
There is a simpler derivation. If we square the whole thing, using the elementary formula (a + b) ^2 = a^2 + b^2 + 2 ab, we get i + -i + 2 sqrt (i) sqrt (-i). The first two terms cancel out, so the result is 2 * sqrt (i * -i), i.e. sqrt (-i^2), i.e. 2. Take the square root of that, and voilà.
(I corrected the earlier version of my previous comment, which was needlessly critical -- this is all for fun, sorry.) There is another equally simple way to see the result. Unfortunately I cannot draw a picture here but if I could I would show the plane with two axes, x and y, and i as [0, 1] in cartesian coordinates and -i as [0, -1] (both of these points are on the y axis). In polar coordinates (distance from the origin, angle from the x axis) they are respectively [1, 90] and [1, -90], expressing angles in degrees. Now the square roots of [ro, theta] are [sr, theta/2] where sr = sqrt (ro) and the symmetric point. So as one of the values for sqrt (i) we get [1, 45] and for sqrt (-i) we get [-1, 45]. (Just divide the angles by 2.) In cartesian coordinates they are [sqrt (2) / 2, sqrt (2) / 2] and [sqrt (2) / 2, -sqrt (2) / 2]. (An isosceles triangle of hypothenuse 1 has sqrt (2) / 2 for each of the other sides -- Pythagoreas's theorem.) If we sum them the y coordinates cancel out and we get 2 * (sqrt (2) / 2), meaning sqrt (2). This is only one of the solutions, we must add symmetry.
Is sqrt(i) * sqrt(-i) = sqrt (i * - i) ? Rules for real numbers don’t necessarily apply to imaginary numbers. Consider the fallacy: -1 = i * i = sqrt[ (-1) * (-1) ] = 1
but with that method any answer is possible as when you multiplied by 2/2 inside the radical you could have chosen any form of a/a. why not choose 4 or some other square number that would be easier to handle. no the original is actually the simplest form.
With all due respect, if taking the square root only gives the positive answer, then if you take the square root of both sides of x^2 = 4, where does the x = -2 come from?
V(i)+V(-i) = V(1,90°)+V(1,-90°) = (1,45°)+(1,-45°) = V2/2(1+i)+V2/2(1-i) = V2 SQRT is a function and has only one value. SO THERE IS ONLY ONE SOLUTION, +V2 Let's give an example ... x^2=4 has 2 solutions : x=2 or x=-2 But there is only one square root of 4 : the square root of 4 is 2, not -2.
If the square root of 4 is 2, not -2, then where did the x = -2 in your example come from? If we take the square root of both sides of x^2=4 and we have to take "ONLY THE ONE SOLUTION", we get x = 2. Your method seems to be missing a solution. In addition, -i can also be (1, 270°) in your notation, so: √(i) + √(-i) = √(1,90°) + √(1, 270°) = (1, 45°) + (1, 135°) = √2(1+i)/2 + √2(-1+i)/2 = i * √2. Please explain if you maintain there's only one solution.
You just think about the two vectors going at +45 and -45 angle, sides are length of 1, forming a triangle of 90 deg, Pythagorean rule, result is sqrt 2. No paper, no calculation, just think of it and you can tell. Btw: depends on how you define sqrt over the body of complex numbers, but how 90% (sic) of mathematicians define it, the result is ONLY + sqrt 2.
I just realized there are so many intelligent peoples in this comment sections👼❤
Yes ,love you all❤
Быдло математику прогуливает, познания гопоты ограничиваются только тем, как бы отнять и поделить. Так что да, тут только культурная братва, остальным делать особо нечего. 😉
You might lose a minus sign in that case.
I would draw a circle in the complex plane and make sqrt(i) and sqrt(-i) into two vectors. I could solve this in my head without pen and paper.
Each root individually has 2 solutions, so there should be 4 solutions. If you map the solutions of root(i) and root(-i) on the complex plain (recall that by taking a root of a complex number you half the phase angle, but you can do this in either direction giving you 2 solutions). And if you then do the addition you'll easily see that all 4 solutions are in a different axis on the complex plain, and that they should be root(2), -root(2), i*root(2) and -i*root(2).
if you didn't write that comment i would have wondered about this the rest of the day
Iagree, if you look at it geometrically, this pops up easily. Just think of i=e^(ipi/2) and the squareroot as dividing that angle into half, so one solution is sqrt(i)=e^(ipi/4) , the other sqrt(i)=e^(i5pi/4). Same with -i, which leads to sqrt(i)=e^(i3pi/4) and sqrt(i)=e^(i7pi/4). Now add them pairwise like vectors, which should lead to all 4 solutions sqrt2, i×sqrt2, -sqrt2 and -i×sqrt2. I would never recommend calculations with complex numbers in only one kind of expression, but switching back and forth and sideways between sum form, exponential form and as geometrical object. By the way, seeing complex numbers as rotation also helps, when multiplied.
@@ralfp8844 Exactly my experience. Algebraic manipulation is the most powerful, but easy to make an error. Oftentimes, geometric representations really helps understanding the problem and breaking it down into easier sub-problems.
Могу сослаться на решение этой задачи в "A Wonderful Math", указанном здесь в первом комментарии. Там представлено решение без возведeния в четвертую степень и тоже только два корня +/- корень из 2 ... .
that would be true, except the radical sign represents the positive solution. if you want both solutions you must write +-✓ . So for this question it is asking for only the positive roots, and so there is 1 solution. However if the question was written as +-(✓i +-✓-i) then there would be four solutions as you say.
I use polar form of complex number for general solution:
i = e^(2m+1/2)πi
sqrt(i) = e^(m+1/4)πi
-i = e^(2n-1/2)πi
sqrt(i) = e^(n-1/4)πi
sqrt(i) + sqrt(-i) = e^(m+1/4)πi + e^(n-1/4)πi = z
Now we have 4 cases to deal with
(m,n) is (even, even): z = e^(π/4)i + e^-(π/4)i = 1/√2 + (1/√2)i + (1/√2 - (1/√2)i) = √2
(m,n) is (even, odd): z = e^(π/4)i + e^(3π/4)i = 1/√2 + (1/√2)i + (-1/√2 + (1/√2)i) = (√2)i
(m,n) is (odd, even): z = e^(-3π/4)i + e^-(π/4)i = -1/√2 - (1/√2)i + (1/√2 - (1/√2)i) = -(√2)i
(m,n) is (odd, odd): z = e^(-3π/4)i + e^(3π/4)i = -1/√2 - (1/√2)i + (-1/√2 + (1/√2)i) = -√2
All four solutions have modulus √2 and either purely real or purely imaginary. We can easily generalize all of them.
sqrt(i) + sqrt(-i) = √2 e^(kπ/2)i where k is any integer.
thanks ❤️
I edited my original post and now my answer matched yours. Things get tricky when not just solving the principal roots.
...for some reason Wolfram Alpha doesn't like the complex solutions though.
Some people argue that sqrt means "principal values only"
I don't know if it is true, if yes, you have to adjust all angles within (-π,π]
√2 is the principal value in this case.
@@mokouf3 that's actually true... Square root is a function.... So there's no way a function can map one value to more than one values...
Actually there are 4 answers, in general
±((√(2)+i√(2))/2)±((√(2)-i√(2))/2)
The main answer is √(2)
But there is also -√(2)
and i√(2)
And also -i√(2)
±√(2) and ±i√(2)
Yeah if call it +-a +-b Rachel takes the solutions as +a+b and -a-b, but +a-b and -a+b should also be solutions.
√2/2[±(1+i)±(1-i)] is cleaner
@@riccardomarino9786 I always forget to factor things, thanks
I don't think so... There are two answers, not 4. Unless I am missing something.
@@bunburyrichard both √i and √(-i) have two solutions. So there are four combinations, and each of them leads to a different result
At 3:00 you lost the ”+/-” between (1+i) and (1-i). So, there are 2 more solutions.
I won't say he "lost" the ± but he made an error when he factored out the ±1/sqrt(2). There is no such thing as a ±1/sqrt(2). It is plus OR minus. Not both. That means there are 4 solutions: a ++ version, a +- version, a -+ version and a - -- version.
On the other hand, I thought his solution was clever. Yeah, there are other ways (square them, convert to exponential functions, etc) but this was fun.
Other than the factorization error.
@@flowingafterglow629 You are of course correct. This may be written as the two expressions
± a (b + c), which is his two solutions and ± a (b - c), which is the missing two. Still 4 solutions, but written in two expressions.
@@flowingafterglow629 I mean ”lost”. That is: not ”lost”, but ” ”lost” ”.😄
I thought there are such an English style to say something like ”+/-sqrt(2)”, just because the solution was too clever to believe it contains such a plain error.
@@flowingafterglow629 The square root radical is defined as being the principal branch **only**.
Thus the correct answer of the problem as posted is single valued at
sqrt(2) * i
evaluated as
(1 + i)/sqrt(2) + (1 - i)/sqrt(2) = (2 + 0i/sqrt(2) = sqrt(2).
Above corrected for a brain-fart error in the original.
The "principal branch" defined for the square root radical is obtained by making a "branch cut" along the negative real axis. This is effected by restricting the value of theta for the original expression (in exponential form) under the radical to being between - pi and + pi.
@@pietergeerkens6324 What is the "principal (sic) branch" of a complex number?
Is exp(i*pi/2) any more "principle" than exp(3*i*pi*2)? Is there a rule that the complex squareroot of a number is the one in the top half of the complex plane?
A shorter way: say A=√i+√-i , then A²=(√i+√-i)²=±2, then A=±√±2, and from here the 4 solutions: √2, i√2,-√2, -i√2.
thank you so much
nice video, i learned new things about complex numbers.
Thanks ❤️❤️
great answer. Another way is to just square the given expression and get i-i-2i^2 =2. Square root this and get +/-sqroot2!
I found it easier to solve using polar form rather than rectangular form complex numbers.
There's only one answer technically if you consider how square roots are defined in the context of complex numbers, the solution with positive real part is chosen while removing the radical... If real part is zero, the one with positive imaginary part is chosen. By chosen, I mean, it becomes the principal square roots.
In general, any given complex number can be expressed in infinitely many ways in the form r•e^(i(x+2nπ)) where n is any integer. So when we apply say square root, the number becomes r•e^(i(x/2+nπ)). We choose n such that x/2 + nπ is an acute angle, (which if not possible, our next preference is obtuse angle or the negative version of acute angle). That's about it... So there's only one square root of a complex number...
√i + √(-i)
= √(e^(iπ/2)) + √(e^(-iπ/2))
= e^(iπ/4) + e^(-iπ/4)
= 2cos(π/4)
= √2
That's the only solution if we consider only the *principal square roots* of the radicals involved.
In practice the "principle" square root is never used. If you are already working with complex numbers, in almost all fields of science you are interested in all roots.
So, which is the cube root of -i?
@@crossiqu they just don’t get it. Hahaha
@@crossiqu if we use the principle cube root of -i, we get that it becomes i. If we instead try to find all solutions to x^3=-i, we have x=i, and x=-i/2±√(3)/2.
-sqrt(2), i*sqrt(2), -I*sqrt(2) are also solutions: it depends on which branch of square root you use. Imagine solutions are when you use different branches for the root.
Great Math Problem
i solved it "geometrically":
visualize i and minus i as arrows in the plain of complex numbers.
i has ankle 90 degrees, -i has ankle 270 degrees.
length of both arrows is 1.
interprete the operation sqrt() as division of ankle by 2.
so you get two arrows, that represent sqrt(i) and sqrt(-i).
add these two arrows as known from vector addition.
you get a triangle which leads you to the solution sqrt(2)*i,
if you apply some basic trigonometry on it.
(in triangle, use: sin(45°)=0.5sqrt(2))
Your method is indeed the polar form. If you talk about all possible solutions, there are 3 more.
If you want principal value only, your solution is not. You need to adjust all initial polar angles to [-π, π] before doing anything.
@@mokouf3
"Your method is indeed the polar form"
yes.
"If you want principal value only...."
i have heard of principal forms, but didn't understand them properly yet. i'll think about it the next days.
how can there be more than one solution to a summation? how about the contradiction that i showed in the proof above?
the contradiction should be wrong if you're right. where exactly is it wrong?
@@KarlHeinzSpock I think that someone shall post a question like this to a forum for public discussion/voting.
@@KarlHeinzSpock There's multiple solution because i has multiple angles. It doesn't have to have a angle of 90. It can also have a angle of 450, or 810, or any angle of the form 360k+90. The polar coordinates (90,1), (450,1), (810,1) all give you the same point on the plane.
@@hrtz9796 all angles of the form 360k+90
deliver only one and always the same number, i.
but sqrt(i) has indeed two solutions, as well as sqrt(-i)
-->the sum sqrt(i)+sqrt(-i) "includes" four different sums, thus has four different solutions.
if you use z = sqrt(i) + sqrt(-i) and square z you will get the result much faster...
thanks
Trebuie sa recunosc ca algoritmul din clip a fost foarte inteligent. Pana la eroare.
According to you, it doesn't give four roots. But it has four roots.
As some astute commenters have pointed out, there are four solutions. The video limits solutions to the real numbers, but the problem involves complex numbers so I think they’re fair game for the solution.
You can derive SQRT(i) by setting it equal to +/- (a + bi) and doing the algebra, but that isn’t necessary. A moment’s thought should bring you to the conclusion that i multiplied by the real roots gives valid solutions because squaring i just flips the sign from +/- to -/+, and both signs are already part of the solution.
Amazing! 👍
√i + √-i =
[into the polar form: re^(iφ), r > 0, -π < φ ≤ π ]
= (e^(iπ/2))^(1/2) + (e^(-iπ/2))^(1/2) = e^(iπ/4) + e^(-iπ/4) =
= 2cos(π/4) = √2
no ± bs, since root functions return a single value
You can not take +/- outside as there were 4 solutions before you did that and now there are only 2, duh 😢
Two results are lost: +sqr(2)*i and -sqr(2)*i.
Wtf, just square the expression and take the root of the result to get back its original value. You’ll get sqrt(2) super easily.
❤️❤️❤️
Technically, that'll give you the numeric value of the result, but it isn't hard to convince ourselves that the correct result must be the positive root.
@@jensknudsen4222 both roots are ok. Maths is a world by itself, and while the “positive” root is sometimes preferred for Analysis and Physiscs, the convention of the positive root being the “correct” one is not as universaly accepted as most people think. If you find this a silly thing, try to take the square root of x=a^2+b^2, and then try to decide wich one is the positive or the negative root.
Such a procedure is always prone to introducing extraneous roots. Eg let x=-1. If you square and square root you get two roots, 1 and -1. Only the second is correct in the setup.
@@ianrobinson8518 1 is also a valid answer, as it works perfectly with the arithmetic done on the setup.
First of all, the square root (√ ) of complex numbers are defined only by the polar form such that the argument becomes halved and the old magnitude is the square of the new magnitude. Eg √(z)=√(r)e^(it/2) given that z=re^(it). This means that the method you use might give the wrong result, but it seems like it does not.
Second, even for real numbers, √(x)>0, so even though the solution for x^2=a has two solutions, √(x) only has one solution for all x>0.
Thirdly, the function √ also only have one solution for the complex domain, where we may only choose the polar representation that gives us 0
First, the complex square root is most commonly defined only for complex numbers whose argument lies in the semi-closed interval (-π, +π] -- in other words, excluding -π, but including +π. Then, halving the argument ensures that the principal square root thus obtained is consistent with real square roots and remains on the same branch for complex numbers. That's why the method works.
Second, √(x)>0 only remains valid for complex numbers if you restrict the domain of x in the way I outlined above. Otherwise, consider the square root of x that your definition produces when x = e^(2πi).
Thirdly, your definition "the function √ also only have one solution for the complex domain, where we may only choose the polar representation that gives us 0
EXCELLENT N VERY NICE
thanks ❤️
The sqrt bracket operator denotes the *principal" square root, hence, no need for considering the negative roots. In other words, it should have been reduced to sqrt(2)*(1 + i + 1 - i) = sqrt(2)*2 = sqrt(2). No need for +/-. If you *do* want to consider non-principal square roots, then you also need to consider the cases where you have opposite selections for the +/- on either root. Hence, sqrt(2)*((1+i)-(1-i)) = sqrt(2)*2*i = sqrt(2)*i. Similarly, another non-principal square root would be -sqrt(2)*i.
A lot of work when you can let it equal answer, A, and square both sides. ONLY TAKES 2 LINES THAT WAY. So A^2 =2 and A = +- rt2
Sorry to nitpick on your solution, but I think you're leaving out an important (though technical) detail. The square root symbol can actually mean two different things:
1. The roots of z^2 = a. For complex a != 0 there are always two roots, one being the negative value of the other. Viewed as a function, this version of the square root is multivalued.
2. The principal value, which is defined as the one of the two roots that has a positive real value (or +i in case the real values are zero). Note that this version is a regular (single valued) function.
Meaning (1) is rarely used, and it's normally pretty safe to assume that the square root symbol refers to the principal value unless otherwise specified.
Assuming this to be the case (that is, both square roots in the given problem both refer to the principal value), the expression only has one value:
sqrt(i) + sqrt(-i) = (sqrt(2)/2)(1 + i) + (sqrt(2)/2)(1 - i) = sqrt(2).
If, on the other hand, if we let the square root symbol denote both roots of the quadratic, then each of the square roots in the original expression has TWO values. Each choice is independent of the other, so we end up with FOUR values:
sqrt(i) + sqrt(-i) = +/-(sqrt(2)/2)(1 + i) +/- (sqrt(2))(1 - i)
where the two +/- operators are independent. Evaluating this all four possible ways yields four distinct values:
sqrt(2), -sqrt(2), sqrt(2) i, -sqrt(2) i.
thanks for your valuable information ❤️❤️
That’s not the only way roots of polynomials work. In fact, the answer depends on the splitting field you want to decompose the polynomial in. If i’m working on Q(sqrt(i);sqrt(-i)) the answer is four-valued, as you state. If i’m working on R(sqrt(i);sqrt(-i)), the answer is +-sqrt(2), as in the video (this one is actually the most common viewpoint), however, if i’m working in C(sqrt(i);sqrt(-i)), then the answer is still sqrt(i)+sqrt(-i). So there are, in general many, many ways to actually “solve” this problem.
@simonflavioibanez7715 Haha... Yes, that's why mathematicians will never be bored. Just when you think you're done, some generalization comes swooping in and knocks over your dominoes.
@@jensknudsen4222 Indeed. xd
FYI: The way to get the "√" symbol on your qwerty keyboard is: "Alt + 2,5,1"
I just visualized the vector for i and halved the argument, and then the same for -i. The imaginary parts cancel, and the real parts add, and there’s your answer.
❤️😊
But did you take the argument for -i to be 3π/2 or -π/2 ? Both are valid arguments for -i, but you get two different complex numbers when you halve the argument. That's a result of the fact that all numbers (except zero), real or complex, have two distinct square roots that are the negative of each other. Had you picked the other argument, you'd find the real parts cancel and the imaginary parts add :o
@@RexxSchneider Every non-zero number does indeed have two square roots, but the radical symbol refers explicitly to the principal square root, which I believe to be defined as the one obtained by starting with -pi < arg(z)
@@MikeGranby If you read the other threads here, you'll find that many commentators don't define the √ to indicate the principal square root when dealing with complex numbers. Others accept that but use a different choice of branch cut such as 0
@@RexxSchneider I can see how one might use a different branch cut, although the one I quoted is the one I was taught, but I think allowing the square radical to be multivalued for is a step too far…
a sum of numbers must not have two different values.
only one of the solutions +-sqrt(2)i is correct. proof:
for example:
0=0
sqrt(i)+sqrt(-i)=sqrt(i)+sqrt(-i)
sqrt(2)i=-sqrt(2)i
2sqrt(2)i=0
this is obviously a contradiction in itself.
only the solution +sqrt(2)i is correct.
i made a mistake. there are four solutions.
You missed 1 possible answer:
i = 1*cis(90)
==>
sqrt(i) = sqrt(1)*cis(90/2 + 360/2 * k), k = 0, 1
so 2 possible answers for sqrt(i):
a= 1*cis(45)
b= 1*cis(225)
-i = 1*cis(270)
==>
sqrt(-i) = sqrt(1)*cis(270/2 + 360/2 * k), k = 0, 1
so 2 possible answers for sqrt(-i):
c= 1*cis(135)
d= 1*cis(315)
Now we can look at all the combinations:
a+c = 0
a+d = sqrt(2)
b+c = -sqrt(2)
b+d = 0
This is all very easy if you look at it in polar form, and addition is just vector addition.
I think:
a+c=sqrt(2i)
b+d=-sqrt(2i)
I find a few problems in the calculation. Sum of two unique values should result in one answer.
If your answer is back substituted
sqrt(i)+sqrt(-i)=+/-sqrt2. Squaring as is
i + 2 sort(i)sqrt(-i) - i =2
2sqrt(i)sqrt(i)sqrt(-1)=2
2i.i=2
2i^2=2
-2=2 so your answer doesn’t satisfy original equation.
isqrt2 would be the correct answer. When you back substitute LHS will simplify the same and the RHS will have an extra -i^2 term making it -2=-2.
By the definition of the the sqrt sign : sqrt(z) is THE principal square root of z, which is unique. For example, sqrt(1)=1 not -1, and sqrt(i)=1/sqrt(2) + i*sqrt(2), Though the equation z^2 = i has two solutions. qrt(-i)=1/sqrt(2) - i*sqrt(2),
The simple and normal solution is by squaring the original equation (gives)... i cancel - i and let 2 multiplied by squared (i) then returning rooting the equation gives =(+&_)root 2
Thanks sir.
we can also solve this using euler's form i=e^(ipi/2) and -i=e^(-ipi/2)
Painful.
Just use and exp(ix) = Cosx + iSinx, and don't forget the +- roots. Then you combine +/-exp(iPi/4) and +/-exp(-iPi/4). You get 4 combinations leading to 2 values: root2 or -root2.
Don't miss out (√2)i and -(√2)i. 4 solutions.
It's not an equation. It is a simple calculation involving functions, which by definition returns only one value.
So there is ONLY ONE VALID ANSWER : +V2
@@tontonbeber4555 If you just talk about principal values
sqrt(i) = e^(π/4)i
sqrt(-i)= e^-(π/4)i
Their sum is indeed √2
But, unlike real functions, a complex function tend to be multi-valued.
@@mokouf3
Try using -i=e^(3Pi/2)
@@boblewis5762 If you talk about principal values only, this one is not used.
If you talk about multi valued version, my general solution included that.
The sqrt symbol means principle value only so don’t use +/-. Also, you made it needle complicated. It’s just 2* Re( sqrt( i) ). And i is exp( i • pi/2 ) so we get 2* Re( exp( i • pi/4 ) = 2* 1/ sqrt( 2 ) = sqrt( 2 ).
Ans: sqrt( 2 ) without +/-.
*needlessly
Also:
sqrt IS a function AND the SOLUTION is ”+/-sqr(...)”.
Like for ”x^2=1” Two real solutions, total 4 solutions.
For a very good explanation, look for Jens Knudsen”s answer to Fred.
Where did you get the "just 2* Re( sqrt( i) )" from? Did you work backwards from what you thought the answer was?
You could also have said that i = exp( i * -3pi/2) so we get 2* Re( exp( i • -3pi/4 ) = 2* (-1/ sqrt( 2)) = -sqrt(2) :o
Just as easy, y = sqr(i)+sqr(-i). y^2=(sqr(i))^2 + 2sqr(i*-i) + (sqr(-i))^2 = i +2 sqr(1) - i = +/- 2
so y = sqr(y^2) = sqr(+/- 2) = +/- sqr(2), +/- i sqr(2)
Nice video but seemed a bit confusing. A beginner would have no idea how to make those clever changes.
Once you write the symbol "√" the choice of sign has already been chosen by you. √4 = 2 not ±2. If you want ±2 then you would need to write ±√4. The square root function is a function. You can only have one output. It maps C to {z in C | Re(z)≥0}. So √i = (1+i)/√2 and not - (1+i)/√2. Similarly √(-i) = (1-i)/√2 . Their sum is √2.
Square root in complex domain has two values, not only one (is not a function)
@@crossiqu What are you talking about? It is a function on the complex plane holomorphic everywhere in C excluding the negative real numbers and 0. The question of this video is about square roots of i and -i, both of which are no where near the negative real axis. Hell, you can take the derivative of it at i and -i. The principal square root, which is universally understood when the √ symbol is written, is well defined.
@@DrR0BERT It's simple. Square root of complex numbers generalized, is not a function in the sense its not return a single value. But you're right for real numbers, and IF you're talking about the PRINCIPAL square root of a complex number. So, OK, the principal square root function is thus defined using the nonpositive real axis as a branch cut. But the written symbol can express both. Its an historic topic regarding n nth ROOTS for nth polinomia. (cube roots and beyond.What are the cube roots of -i?) So the formula it's a kind of ambigous
@@crossiqu What are the cube roots of -i? There are three of them. That's not the issue. Nor is the issue that both i and -i have two square roots. The issue is when you use the symbol √ you are identifying only one of them, namely the principal square root. If you want to reference the non-principal square root, you would use -√ .
Let me make this simple. The two questions: "What is √(-i)?" and "What are the square roots of i?" do not have the exact same answer. The first one is a set of one number, and the second is a set of two.
@@DrR0BERT You may find it helpful to understand that using the surd symbol √ to mean the principal square root of a number has not always been the convention. You will find that different audiences of different nationalities and different text books may utilise conventions in a way that differs from yours. Admittedly, you've probably got to examine text books written in English that are over 100 years old to see √ used to mean both square roots, and I agree that the majority of serious mathematicians working in English today would expect the symbol √ to denote the principal square root, but I don't think it's helpful to simply dismiss alternative interpretations as invalid.
I got the answer by squaring the expression and finding the roots +/- sqrt(2).
Good video. 👍
thanks ❤️❤️
Let the single letter p be pi, sqrt(i) + sqrt(-i) = cis(p/4)+cis(-p/4), vector1= 45 degree (positive x axis) with a length of 1 from origin +vector2= -45 degree with a length of 1 from origin, give you the resultant vector of sqrt(2) length lying on the positive X axis, so answer is real number sqrt(2) [Right angle triangle]
There is only one value for this, the square root is not a multi-valued function by definition, and your insertion of a +/- is unambiguously wrong. This solution is not only overly long (compared to using the exponential definition) but also incorrect.
thanks ❤️❤️
Awesome ❤❤
I agree with sqrt(I)=(1+i)/sqrt2
sqrt(-i)=sqrt(-1)*sqrt(i)=i(1+i)/sqrt2=(-1+i)/sqrt2
So sqrt(i)+sqrt(-i)=2i/sqrt2
Using polar form you get the same result. This answer satisfies original equation.
Thanks
Brillant
Assuming that you are taking the surd symbol √ to indicate the principal square root of the complex number, we proceed like this:
Write the complex number in polar form, so i = e^(πi/2). That has two square roots, e^(πi/4) and e^(πi/4 + πi).
The principal square root of a complex number r.e^(iθ) whose argument (θ) satisfies -π < θ
thanks
Great 👍
i looked up some basics.
let C be the set of all complex numbers.
the operation + of two complex numbers is then defined as a function
'+': (CxC)-->C. this is part of a group theory axiom.
since '+' is a function, the sum x+y of two complex numbers must have exactly one solution.
so only +i*sqrt(2) is a solution.
proove me wrong, if you know something that i don't know or that i misunderstood, since i'm not completely shure about what i said above.
it's not about who knows best, it's about learning something.
i'm on the trail of my own misunderstandings:
sqrt(i) is not only one number, there are two numbers coming out of sqrt(i).
it can have the angles 45° and 225°
sqrt(-i) can result in an angle of 135° and an angle of 315°
so sqrt(i)+ sqrt(-i) represent four different additions with four differe values.....
formally, x+y here can be denoted as a solution z of a set of equations: z = x + y, x^2 = i, y^2 = -i
@@88coolv
thanks for your answer.
what you said is exactly what i learned about complex numbers while thinking about this problem.
To do it the way you did it, you kind of have to anticipate the answer. Why not use 3/3 or 4/4 instead of 2/2? Using 2/2 allows you to get to (1+i)^2 and (1-i)^2 and solve it the way you did. The answers in the comments that did it geometrically or by squaring both sides showed me why sqrt(2) is correct.
Why should we adopt such a lengthy process when short-cut is available?
Just use i and -I exponential notation you get e^pi/4+e-pi/4 by simply fine using the square roots and the properties of exponential but this is 2cos(pi/4) using Euler’s trigonometric identify which is equal to sqrt(2) finally
thanks
+/- sqrt(a) +/- sqrt(b)
Can also be:
sqrt(a) - sqrt(b)
&
-sqrt(a) +sqrt(b)
i^(1/2) + (-i)^(1/2)
=
e^(i x pi / 2 + 2 x pi x m)^(1 / 2)
+
e^((-i) x pi / 2 + 2 x pi x n)^(1 / 2)
=
e^(i x pi / 4 + pi x m)
+
e^((-i) x pi / 4 + pi x n)
=
(e^pi)^m x e^(i x pi / 4)
+
(e^pi)^n x e^(-i x pi / 4)
Now without the m:
e^(i x pi / 4) =sqrt(2)/2 + sqrt(2)/2
& without the n:
e^(-i x pi / 4) =sqrt(2)/2
-
i x sqrt(2)/2
If m=n=+/- 2 then you can to this.
But if m= 2 & n = -2 you have another solution.
And with m=-2 & n=2 yet another.
It's not difficult to see there are other solutions .
For all odd m & n.
No just easily
We calculate the expression and we square it (a+b)² =a²+b²+2ab
We find 2 so ±2
Best Teaching 💙
Wouldn't it be easier to square the whole thing, then +i and -i cancel out and in the "2*sqrt(i * -i)" part the square root evaluates to 1, leaving just 2. Take a square root of that and add +/- in front of it.
👍👍👍👍👍👍
In this kind of a problem you have to add the principal square roots. Principal square root of -i is (i-1)/sqrt2
So your answer would be isqrt2.
If you used polar form the problem wouldn’t be there.
I=e^PI/2 so sqrt (I)=e^PI/4 = (1+i)/sqrt2
-I=e^3PI/2 so sqrt(-I)=e^3PI/4=(-1+i)/sqrt2
When you add the two you get the same answer I quoted.
Why not use absolute values in the end. Which is the same as modulus of (1 + i) which is square root of 2.
On a polar coordinate system the principal root is the first one you find counter clockwise. When you add those ones you always get the same answer.
I have simple solution for above equation, may be required correction:
Let √i + √-i = x
squaring on both side,
(√i + √-i)² = x²
Solving LHS,
(√i + √-i)² = (√i)² + 2(√i)(√-i) + (√-i)²
= i + 2√(i)(-i) -i
=2√-(i²)
=2√-(-1)
=±2
After simplifying the left-hand side, we have (√i + √-i)² = ±2.
therefore x² = ±2
x=±√±2
solution are = +√+2, +√-2, -√+2, -√-2
= √2, i√2, -√2, -i√2
√i + √-i = ±√2, ±i√2
thanks my friend ❤️❤️
You can find out it nicely using argand diagram.
Very lengthy process followed.
Just take square of the whole expression and expand. Then if it is equal to square of 'x', then retake the square root of the simplified result to get value of 'x' i.e. the given expression.
I meant simplified end result of the expanded term as mentioned.
Is it allowd to thake the sqrt of a negativ imaginary number in the complex plane or do you need more dimantions for that?
It is allowed. Any negative imaginary number can be written as r * e^(-π/2) where r is a positive real number. Its principal square root is normally defined to be √r * e^(-π/4), taking the positive value of the square root of r. Its other square root is √r * e^(3π/4).
طريقة اسهل نفرض a= (الجزر التربيعي ل i+الجذر التربيعي ل i_) نحسب aتربيع ونبدل عن قيمةiمربع ب1_فنجد المطلوب
Sir: could we not express i and -i in terms of e^(ix) and then change simplify in terms of cos and sin?
Geometrically, you end up with 4 vectors 90 deg apart and splitting each quadrant in half. Adding vectors geometrically, you are finding the diagonals of 4 unit squares each of which lie on the R axis or the I axis. Of course, the diagonal of a unit square = sqrt(2).
😊
How does the sum of 2 roots equals 2 numbers?
If you're familiar with the properties of complex numbers this problem can be solved by inspection.
By writing i in polar form one finds sqrt(i) + sqrt(-i) = +/- 2·cos(π/4) = +/- sqrt(2) ◼
No complex algebra required.
You've assumed that the ± i.sin(π/4) cancels. But the ± is independent for the case of sqrt(i) and sqrt(-i), so we find that in the four resulting cases, two of them cancel the imaginary parts, giving ±sqrt(2), but the other two cases cancel the real parts, giving ±i.sqrt(2).
@@RexxSchneider Pardon me for pointing it out, but you're wrong. i = e^(π i/2 + 2 π k i), where k is an integer. Therefore, sqrt(i) = e^(π i/4 + π k i). It's obvious then that sqrt(- i) is the conjugate of sqrt(i), and k is in the set {0,1}. When k = 0, sqrt(i) = cos(π/4) + i sin(pi/4); sqrt(-i) = cos(pi/4) - i sin(π/4). Summing both expressions together yields 2·cos(π/4). When k
= 1, sqrt(i) = - cos(π/4) - i sin(π/4); likewise, sqrt(-i) = - cos(π/4) + i sin(pi/4). Again summing these two expressions yields - 2·cos(π/4). These are the only values of the original expression sqrt(i) + sqrt(-i), as assigning k any value greater than 1 or less than 0 causes the cycle to repeat, yielding the same two possible answers: +/- 2·cos(π/4) ◼
@@johnnolen8338 I'm also sorry to point it out, but you're wrong. The k in the polar form you write for i and the k in the polar form you write for -i are independent.
You could call them k1 and k2 if you wish, and you're quite right that they both lie in the reduced set { 0, 1 }.
But you've missed the case where k1 = 0 and k2 = 1 and the case where k1 = 1 and k2 = 0.
Try those out and see which parts cancel out. Did you find the other two possibilities, now?
@@RexxSchneider I'm not going to debate this; k is an integer. As such you only have to include it once when defining i in polar form. -i is the complex conjugate of i. Therefore the k's are not independent. They have the same absolute value but opposite signs. Whenever you add a complex number to its complex conjugate the sum is always twice the real part of the complex number. The imaginary parts cancel because they have opposite signs. If k is even you get the positive value else if k is odd,you get the negative value. There are no other possibilities. Don't believe me? Plot it on the unit circle for yourself. But in the meantime stop playing stump the dummy with people who are way more experienced than you.
@@johnnolen8338 There's nothing to debate. If you can't see that there are two numbers, and each of them can have *any* integer in its polar representation., then you're too stupid to bother explaining things to.
Your k's are clearly independent. If we were evaluating √(ai) + √(bi) would you still insist that √(a.e^πi(2m + 1/2) + √(b.e^πi(2n + 1/2) had to have n=m? Of course not, there's no such restriction. Then ask yourself what's magical about a=1; b=-1 that suddenly forces n to be equal to m?
You state "Whenever you add a complex number to its complex conjugate the sum is always twice the real part of the complex number." Yes, of course. We know that (a+ib) + (a-ib) = 2a. But we're dealing with √(a+ib) and √(a-ib) and we can't are just call them "conjugates".
The square root of i has two distinct values, A = (1+i)/√2 and B = (-1-i)/√2.
The square root if -i has two other distinct values, C = (1-i)/√2 and D = (-1+i)/√2.
You get four distinct possible sums when you add them together, A+C, A+D, B+C, B+D.
You can't just decide to ignore two of those.
Work it out and see how much you've embarrassed yourself.
Of course I don't believe you. I'm a 71-year-old Cambridge graduate who spent a lifetime teaching maths, and you're an ignorant child who is unlikely to ever match my experience if you live another 100 years.
How could you consider √[(1 - i)^2] = ± (1 - i) ???
Is (1 - i) > 0 or (1 - i) < 0 ???
1 > i or i < 1
Let it=t, then it's very easy to identify t^2=i-i+2sqrt(i*(-i))=2
It's helpful for students
❤️❤️
it's positive sqrt from the beginning, why do you put "plus minius"?
You write i^2-2.1.i+i^2. as (1-i)^2. and then use that ⎷(1-i)^2. equals 1-i
What if you write i^2-2.1.i+i^2. as (i-1)^2. Then you would have found ⎷(i-1^2. equals i-1. Or not? The problem is that when you have
⎷(1-i)^2 and replace it by 1-i. you have to find out first if 1-i or i-1 is the principal root of (1-i)^2. Because ⎷ stand for the principal root of a value. And then in the situation of (i-1)^2 we also have ⎷(i-1^2. equals 1-i.
Just using the fact that ⎷(1-i)^2. equals 1-i and not i-1 without an explanation makes the solution incomplete.
Two answers...that is great, but n general a square root of a complex number remains multi valued so in that sense the question is "worded" in a non conform way. It would be better to call the given expression x and then through squaring obtain a polynomial equation and then find all the roots. Equally, the assumption i = sqrt(-1) is incorrect, it should be "i is the imaginary unit for which its square is -1"
You cannot factor out "±", you have missed some solutions there.
The notation sqrt(i) is a problem no (sqrt only for positive numbers) ?
I mean how the same notation means two different numbers ?
As I - not always, but often - do when I watch math videos, I stopped the video before watching the solution, took paper and pen and tried it on my own.
I got the result i*sqrt(2).
Then I continued watching the video to compare. After that I was surprised to see sqrt(2).
At first I thought I had done something wrong, but after reading the comments I was confirmed, that my result was correct as well.
So, all in all, there are 4 different possible solutions.
- Interesting question, that was fun!
😊😊
I did this by squaring the binomial and it simplified to 2, and the square root comes out to the square root of 2.
That's how I did it too.
While that works in this case I'm not sure that is valid in general.
For anything that summed to a number with negative real part it would be wrong (just (-0.5)+(-0.5) would fail it giving +1) .
Sir apner to 1 lakh view aice🎉🎉🎉🎉
Do this in polar coordinates: Show the geometric approach!
I humbly disagree with everyone here asserting that there are four "solutions" to this problem. As this is presented, we are not being asked to solve for anything. We are simply asked to simplify the given expression. If we were instead given the expression (√4 + √9), one would simply evaluate the principle root of each term, and give the answer 5. At least as I was always taught, -5, -1, and 1 are not "alternative" or further correct answers. However, if we had been given two quadratic equations, x=√(4) and y=√(9), and asked to evaluate the sum x+y, then our solution set would consist of all four elements. The fact that this problem involves complex numbers seems to me irrelevant to this general point. Here we are only evaluating an expression, so we sum the two principle roots. Hence, √2*i
At least that's what my math teachers from middle school through college told me. 🤷
to me (french, master degree in electronics) the sqare root function is only defined from R+ to R+. There is no sqrt(i) in my math. Nowhere to be seen. Only complex number x so that x^2=i (or any other complex number and any number of powers of x). I had never seen
I think what many commentators are saying is that they can see how √i + √(-i) could be evaluated to yield four different values. If you have been taught -- as the majority have -- that functions have only one value, that therefore we have to choose a "principal value" from the two possible square roots of any number, and that the surd symbol √ denotes the principal square root, then you will only find one value for √i + √(-i), namely √2. Nevertheless, I think from the volume of commentators who take a different approach, namely that √ can represent either the principal square root or its negative, that your understanding does not encompass the entire field of possibilities, and it is instructive to understand other views.
There is a simplistic alternative solution..consider the whole expression as x and square that. x*x=2, x,=+/-sq rt of 2
As far as I know, this term is unsovable, because negative numbers don´t have a square root. If "i" is positive, then "-i" must be negative and vice versa. If I am wrong, please tell me why. Thanks a lot.
your answer is correct, if you refer to the set of real numbers, those "normal" numbers known from school maths.
the given math problem refers to a extended set of numbers, named complex numbers. complex numbers allow square roots of negative numbers.
if you want to understan it, read first about complex numbers and their properties.
i can't explain it here, it is too much stuff.
@@KarlHeinzSpock thank you.
From the polar form,
sqrt(i) = +/- [sqrt(2)/2 + i sqrt(2)/2]
sqrt(-i) = +/- [sqrt(2)/2 - i sqrt(2)/2]
sqrt(i) + sqrt(-i) = +/- sqrt(2).
It was fun seing him do it the hard way 🤣
√i +√-i= e^iπ/4 +e^-iπ/4 =2.cos(π/4)=√2.
Nice🎉 Who needs to learn maths easy and fast? Check it😊
thanks ❤️❤️❤️❤️❤️❤️
Если обе стороны уравнения возвести в квадрат, x²=(√i+√(-i))². Тогда:
▫x²=i+2√i√(-i)-i;
▫x²=2√i√(-i);
▫√a√b=√(ab);
▫√(i*(-i))=√(-i²);
▫i²=-1;
▫-(-1)=1;
▫√1=1;
▫x²=2*1;
▫x²=2;
▫x=±√2.
Thanks❤️❤️
as said you get 4 possibilitis (+ - ) + ( + -) with 2 different solutions sqrt( 2) 0r sqrt( 2) * j
There is a simpler derivation. If we square the whole thing, using the elementary formula (a + b) ^2 = a^2 + b^2 + 2 ab, we get i + -i + 2 sqrt (i) sqrt (-i). The first two terms cancel out, so the result is 2 * sqrt (i * -i), i.e. sqrt (-i^2), i.e. 2. Take the square root of that, and voilà.
(I corrected the earlier version of my previous comment, which was needlessly critical -- this is all for fun, sorry.) There is another equally simple way to see the result. Unfortunately I cannot draw a picture here but if I could I would show the plane with two axes, x and y, and i as [0, 1] in cartesian coordinates and -i as [0, -1] (both of these points are on the y axis). In polar coordinates (distance from the origin, angle from the x axis) they are respectively [1, 90] and [1, -90], expressing angles in degrees. Now the square roots of [ro, theta] are [sr, theta/2] where sr = sqrt (ro) and the symmetric point. So as one of the values for sqrt (i) we get [1, 45] and for sqrt (-i) we get [-1, 45]. (Just divide the angles by 2.) In cartesian coordinates they are [sqrt (2) / 2, sqrt (2) / 2] and [sqrt (2) / 2, -sqrt (2) / 2]. (An isosceles triangle of hypothenuse 1 has sqrt (2) / 2 for each of the other sides -- Pythagoreas's theorem.) If we sum them the y coordinates cancel out and we get 2 * (sqrt (2) / 2), meaning sqrt (2). This is only one of the solutions, we must add symmetry.
Is sqrt(i) * sqrt(-i) = sqrt (i * - i) ? Rules for real numbers don’t necessarily apply to imaginary numbers. Consider the fallacy: -1 = i * i = sqrt[ (-1) * (-1) ] = 1
(√i+√-i)^2=i+(=i)+2√-i^2
= 2√1
So taking square root on both sides we get
√i+√-i= + or - √2
but with that method any answer is possible as when you multiplied by 2/2 inside the radical you could have chosen any form of a/a. why not choose 4 or some other square number that would be easier to handle. no the original is actually the simplest form.
+-√2...... thirty second job 😊😊
I think using the polar method is best.
sqrt(i)(1+i)
You lost two solutions on 3:00
thanks
I just used (a+b)^2=a^2+2ab+b^2 and then it cancels out etc.....
With respect, there should only be the positive answer. Square root of 4 is 2. However, if x^2 =4, then x=+- 2.
With all due respect, if taking the square root only gives the positive answer, then if you take the square root of both sides of x^2 = 4, where does the x = -2 come from?
V(i)+V(-i) = V(1,90°)+V(1,-90°) = (1,45°)+(1,-45°) = V2/2(1+i)+V2/2(1-i) = V2
SQRT is a function and has only one value.
SO THERE IS ONLY ONE SOLUTION, +V2
Let's give an example ... x^2=4 has 2 solutions : x=2 or x=-2
But there is only one square root of 4 : the square root of 4 is 2, not -2.
If the square root of 4 is 2, not -2, then where did the x = -2 in your example come from?
If we take the square root of both sides of x^2=4 and we have to take "ONLY THE ONE SOLUTION", we get x = 2. Your method seems to be missing a solution.
In addition, -i can also be (1, 270°) in your notation, so:
√(i) + √(-i) = √(1,90°) + √(1, 270°) = (1, 45°) + (1, 135°) = √2(1+i)/2 + √2(-1+i)/2 = i * √2. Please explain if you maintain there's only one solution.
Just write i as _/-1 and you'll get your answer
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That will just make it more complex
You just think about the two vectors going at +45 and -45 angle, sides are length of 1, forming a triangle of 90 deg, Pythagorean rule, result is sqrt 2. No paper, no calculation, just think of it and you can tell.
Btw: depends on how you define sqrt over the body of complex numbers, but how 90% (sic) of mathematicians define it, the result is ONLY + sqrt 2.
The answer is sqrt(2), not +/- sqrt(2)
just compute (sqr(i)+sqr(-i))^2 = i-2-i = 2 so the result is +/- sqr(2)
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